## The Elements of Euclid: Viz. the First Six Books, Together with the Eleventh and Twelfth |

### From inside the book

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Page 17

1 . right angles to AB ; and because ABC is a straight line , the angle CBE is equal * to the angle EBA ; in E the same

1 . right angles to AB ; and because ABC is a straight line , the angle CBE is equal * to the angle EBA ; in E the same

**manner**, because ABD is a straight line , the angle DBE is equal to ... Page 20

And in like

And in like

**manner**, it may be demonstrated that no other can be in the same straight line with it but BD , which therefore is in the same straight line with CB . Wherefore , if at a point , & c . Q. E. D. PROP . XV . THEOR . Page 22

In like

In like

**manner**, it may be demonstrated , that BAC , ACB , as also CÁB , ABC , are less than two right angles . Therefore any two angles , & c . Q. E. D. angle BĆA PROP . XVIII . THEOR . The greater side of every triangle is opposite lo ... Page 23

In the same

In the same

**manner**it may be Because AD is equal t to AC , add BA to each , therefore + Conse the whole BD is equal t to the two BA and AC . * 3.1 . 5.1 . D far of When Deter San Take a 9 demonstrated. 12 Ax . * 3 . 1 . * 4 . Page 30

... but it + Hyp . is also equals to it , which is impossible ; therefore AB A E B and CD being produced do G not meet towards B , D . In CF like

... but it + Hyp . is also equals to it , which is impossible ; therefore AB A E B and CD being produced do G not meet towards B , D . In CF like

**manner**it may be demonstrated , that they do not meet towards A , C : but those straight ...### What people are saying - Write a review

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### Other editions - View all

The Elements of Euclid: Viz. The First Six Books, Together With the Eleventh ... Euclid Euclid No preview available - 2018 |

### Common terms and phrases

altitude angle ABC angle BAC base base BC BC is equal centre circle ABCD circumference common cone Const contained cylinder demonstrated described diameter divided double draw drawn equal angles equiangular equilateral equimultiples extremities fall figure fore four fourth given given straight line greater half inscribed join less Let ABC likewise magnitudes manner meet multiple opposite parallel parallelogram pass perpendicular plane polygon prism PROB produced PROP proportionals proved pyramid ratio reason rectangle contained rectilineal figure remaining angle right angles segment shown sides similar solid solid angle solid parallelopiped sphere square taken THEOR third touches triangle ABC vertex wherefore whole